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I originally thought C, but I changed my mind. I think it's A. I thought originally that we had to be sure they weren't 0 or 1, because either of those factorial make 1, and 1 is odd. But knowing that the variables are equal means that even if they are 0 or 1, they both make 1, and 1 + 1 is even. So that's enough.
In 2, knowing that one variable is 2 doesn't tell us about the other one, which could be 0 or 1. So in either case, we'd get 3, which is odd, but in every other case we'd have even plus even which is even (every number other than 0 or 1 factorial must be even since it has at a minimum 2 times something in it).
I originally thought C, but I changed my mind. I think it's A. I thought originally that we had to be sure they weren't 0 or 1, because either of those factorial make 1, and 1 is odd. But knowing that the variables are equal means that even if they are 0 or 1, they both make 1, and 1 + 1 is even. So that's enough.
In 2, knowing that one variable is 2 doesn't tell us about the other one, which could be 0 or 1. So in either case, we'd get 3, which is odd, but in every other case we'd have even plus even which is even (every number other than 0 or 1 factorial must be even since it has at a minimum 2 times something in it).
Infact, from the stem, it looks like it will never be an odd number. You will never get 0! or 1! from E!+Q!
Besides 1! factorial every other factorial is multiplied by atleast 2 an even number and any integer multiplied by an even number will always be even.
yeah. in fact, I think it would be even no matter what. I didn't see the exclamation point after the absolute value signs originally. I saw it without that, so there was the chance of getting 2+1=3.
I will go with D.
At first i missed the outside factorial sign the moment i saw i knew all number greater than 2 will give even number. So even the factorials are O or 1 you have 2.